Question

The null and alternative hypotheses are:

H0:μd≤0H0:μd≤0

H1:μd>0H1:μd>0

The following sample information shows the number of defective units produced on the day shift and the afternoon shift for a sample of four days last month.

	Day
	1	2	3	4
Day shift	11	10	14	19
Afternoon shift	10	9	14	16

At the 0.01 significance level, can we conclude there are more defects produced on the Afternoon shift?

a. State the decision rule. (Round the final answer to 3 decimal places.)

H₀ is (rejected or not rejected?) if t > _______

b. Compute the value of the test statistic. (Round the final answer to 3 decimal places.)

Test statistic: ___________

c. What is your decision regarding the null hypothesis?

H₀ should (be rejected or not be rejected?)  

d. Determine the p-value. (Round the final answer to 4 decimal places.)

The p-value is ______

Answer 1

a.

Let d be the difference between defects produced on the Afternoon shift and defects produced on the Day shift.

The null and alternative hypotheses are:

H0: μd ≤ 0

H1: μd > 0

Degree of freedom = 4-1 = 3

Critical value of t at df = 3 and 0.01 significance level is 4.541

H0 is rejected if t > 4.541

b.

d = 10-11 , 9-10, 14-14, 16 - 19

= -1, -1, 0, -3

Sample mean, = (-1 - 1 + 0 - 3)/4 = -1.25

Sample standard deviation, s = sqrt[((-1 + 1.25)^2 + (-1 + 1.25)^2 + (0 + 1.25)^2 + (-3 -+1.25)^2 )/(4-1)] = 1.258306

Standard error of mean difference, SE = s / = 1.258306 / = 0.629153

Test statistic, SE = / SE = -1.25 / 0.629153 = -1.987

c.

H0 should not be rejected as t < 4.541

d.

P-value = P(t > -1.987) = 0.9295